When a surface 1 cm thick is illuminated with light of wavelength λ the stopping potential is V0 but when the same surface is illuminated by light of wavelength 3λ the stopping potential is V0 /6. the threshold wavelength for metallic surface is

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4λ

5λ

3λ

2λ

answer is B.

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Detailed Solution

Here eV0=hc1λ−1λ0and eV06=hc13λ−1λ0 Dividing eq. (1) by eq. (2), we get 6=1λ−1λ013λ−1λ0 or 63λ−6λ0=1λ−1λ0 or 63λ−1λ=6λ0−1λ0 or 1λ=5λ0 or λ=λ05 or λ0=5λ

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